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musickatia [10]
3 years ago
8

g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p

erpendicularly to the field. Under these conditions, the electron undergoes circular motion. Find the radius r of the electron's path and the frequency f of the motion.
Physics
2 answers:
grin007 [14]3 years ago
6 0
<h2>Answer:</h2><h2></h2>

(a) 422.2 x 10⁻⁷m

(b) 4.6 x 10⁵ Hz

<h2>Explanation:</h2>

The magnetic force acting perpendicular to the velocity of the electron will cause a circular motion so that the centripetal force, F_{c}, of the electron is equal to the Lorentz's force, F_{l}. i.e

F_{c} = F_{l}         ---------------(i)

Where;

F_{c} = \frac{mv^2}{r}

F_{l} = qvB

Equation (i) then becomes;

\frac{mv^2}{r} = qvB           ------------------(ii)

Where;

m = mass of the electron

v = linear velocity of the electron

r = radius of the electron's path

q = charge of the electron

B = magnetic field.

Make r subject of the formula in equation (ii)

r = \frac{mv^2}{qvB}

r = \frac{mv}{qB}                     -----------(iii)

From the question;

v = 121m/s

B = 1.63 x 10⁻⁵ T

q = 1.6 x 10⁻¹⁹ C             (known constant)

m = 9.1 x 10⁻³¹kg

Substitute these values into equation (iii) as follows;

r = \frac{9.1*10^{-31} * 121}{1.6*10^{-19}*1.63*10^{-5}}

r = 422.2 x 10⁻⁷m

Therefore, the radius of the electron's path is 422.2 x 10⁻⁷m

(ii) The frequency, f, of the motion which is also called the cyclotron frequency - the number of cycles the electron completes around the path every second - is given by;

f = \frac{v}{2\pi r}

Substitute the values of v and r into the equation as follows;

f = \frac{121}{2\pi (422.2*10^{-7})}

Take \pi = 3.142

f = \frac{121}{2(3.142) (422.2*10^{-7})}

f = 4.6 x 10⁵ Hz

Therefore, the frequency of the motion is 4.6 x 10⁵ Hz

kolbaska11 [484]3 years ago
3 0

Answer:

i. The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

                 r = \frac{mv}{qB}

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × 10^{-5}T, v = 121 m/s, Θ = 90^{0} (since it enters perpendicularly to the field), q = e  = 1.6 × 10^{-19}C and m = 9.11 × 10^{-31}Kg.

Thus,

         r = \frac{mv}{qB} ÷ sinΘ

But,  sinΘ =  sin 90^{0} = 1.

So that;

          r = \frac{mv}{qB}

            = (9.11 × 10^{-31} × 121) ÷ (1.6 × 10^{-19}  × 1.63 × 10^{-5})

            = 1.10231 × 10^{-28}   ÷ 2.608 × 10^{-24}

            = 4.2266 × 10^{-5}

            = 4.23 × 10^{-5} m

The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

B. The frequency 'f' of the motion is called cyclotron frequency;

           f = \frac{qB}{2\pi m}

             =  (1.6 × 10^{-19}  × 1.63 × 10^{-5}) ÷ (2 ×\frac{22}{7} × 9.11 × 10^{-31})

             =  2.608 × 10^{-24} ÷  5.7263 × 10^{-30}

             = 455442.4323

          f  = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

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